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FAO David Broadhurst, Brun sum

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  • Guy
    Hi David I m afraid I haven t the time to do this properly, (life, grr) but it appears you can write a modified Brun s sum like this C_2(X)/log x +
    Message 1 of 1 , Nov 6, 2012
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      Hi David
      I'm afraid I haven't the time to do this properly, (life, grr) but it appears you can write a modified Brun's sum like this

      C_2(X)/log x + m*integral(3,X) C_2(t)dt/(t*log^2 t)

      where C_2(X) is sum(p twin prime>2 log p/p) (ie log(3)3+log(5)/5+log(11)/11+log(17)/17+log(29)/29 +...)

      and m=1 (this then=1.058... like we discussed ten years ago)

      now what's interesting is you can choose m to be less than one, and the limit then appears to be O(log^-2 X), as X->inf. for certain m,say m=1/(4*C_twin) and C_twin=.66016...

      sorry I can't explain better, life gets in the way all the time.
      from Guy
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